nth moments using moment generating functions

**FGT12** · May 29th 2011, 08:15 AM

For a standard random variable Z let $\mu_n = E[Z^n]$
Show that:

$\mu_n = 0$ when n is odd
$\mu_n = \frac{(2j)!}{2^jj!} \when\ n=2j$

I started:
$E[e^tZ]=e^{{t^{2}}/2}$ from moment generating function of a normal variable
$= \displaystyle\sum_{j=0}^\infty \frac{(t^2/2)^j}{j!}$
$= \displaystyle\sum_{j=0}^\infty \frac{t^{2j}}{2^jj!}$

however I cannot see where to go now

**TheEmptySet** · May 29th 2011, 08:43 AM

Originally Posted by FGT12

For a standard random variable Z let $\mu_n = E[Z^n]$
Show that:

$\mu_n = 0$ when n is odd
$\mu_n = \frac{(2j)!}{2^jj!} \when\ n=2j$

I started:
$E[e^tZ]=e^{{t^{2}}/2}$ from moment generating function of a normal variable
$= \displaystyle\sum_{j=0}^\infty \frac{(t^2/2)^j}{j!}$
$= \displaystyle\sum_{j=0}^\infty \frac{t^{2j}}{2^jj!}$

however I cannot see where to go now

I assume you mean a standard normal random variable then

$\mu_n=\mathbb{E}[z^n]=\frac{1}{2\pi \sigma^2}\int_{-\infty}^{\infty}z^ne^{\frac{-(z-\mu)^2}{2\sigma^2}}$

The Gaussian is an even function and if $\mu=0$ then

$z^n$

Is even or odd depending on the exponent.

So what do you know about integrals about even and odd functions over symmetric intervals?

**matheagle** · May 29th 2011, 10:07 PM

what is j in this???

$\mu_n = 0$ when n is odd
$\mu_n = \frac{(2j)!}{2^jj!} \when\ n=2j$

Ok the math killed the english, I see it when I quote you.

$\mu_n = \frac{(2j)!}{2^jj!}$ when n=2j

why not just write

$\mu_n = \frac{n!}{2^{n/2}(n/2)!}$ if n is even

Did you try differentiating the MGF?

**matheagle** · May 29th 2011, 10:20 PM

I would just use the gamma function

$E(Z^n)={1\over \sqrt{2\pi}}\int_{-\infty}^{\infty}z^ne^{-z^2/2}dz$

$=\sqrt{2\over \pi}\int_0^{\infty}z^ne^{-z^2/2}dz$

Then let $w=z^2/2$

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